Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2}{x + 7} = \dfrac{-x + 42}{x + 7}$
Multiply both sides by $x + 7$ $ \dfrac{x^2}{x + 7} (x + 7) = \dfrac{-x + 42}{x + 7} (x + 7)$ $ x^2 = -x + 42$ Subtract $-x + 42$ from both sides: $ x^2 - (-x + 42) = -x + 42 - (-x + 42)$ $ x^2 + x - 42 = 0$ Factor the expression: $ (x - 6)(x + 7) = 0$ Therefore $x = 6$ or $x = -7$ However, the original expression is undefined when $x = -7$. Therefore, the only solution is $x = 6$.